Q.(a) Show that the matrices $A = \begin{pmatrix} 2 & 2 & 1 \\ 1 & 3 & 1 \\ 1 & 2 & 2 \end{pmatrix}$ and $B = \begin{pmatrix} \frac{4}{5} & \frac{-2}{5} & \frac{-1}{5} \\ \frac{-1}{5} & \frac{3}{5} & \frac{-1}{5} \\ \frac{-1}{5} & \frac{-2}{5} & \frac{4}{5} \end{pmatrix}$ are inverse of each other.

(a) Multiplying $A$ by $B$ gives the identity matrix, proving they are inverses. (b) $\tan 2\alpha = \tfrac34$; then $\tan(2\alpha+\beta) = 1$, so $2\alpha+\beta = \tfrac{\pi}{4}$.

Part (a): Show $A$ and $B$ are inverses.

Two square matrices are inverses iff their product is the identity $I$. Compute $AB$ (with $B = \tfrac15\begin{pmatrix}4&-2&-1\\-1&3&-1\\-1&-2&4\end{pmatrix}$).

• Row 1 $\times$ B: $\tfrac15(2\cdot4 + 2\cdot(-1) + 1\cdot(-1)) = \tfrac{5}{5}=1$; $\tfrac15(2\cdot(-2)+2\cdot3+1\cdot(-2)) = \tfrac{0}{5}=0$; $\tfrac15(2\cdot(-1)+2\cdot(-1)+1\cdot4) = \tfrac{0}{5}=0$.
• Row 2 $\times$ B: $\tfrac15(1\cdot4+3\cdot(-1)+1\cdot(-1)) = 0$; $\tfrac15(1\cdot(-2)+3\cdot3+1\cdot(-2)) = \tfrac{5}{5}=1$; $\tfrac15(1\cdot(-1)+3\cdot(-1)+1\cdot4)=0$.
• Row 3 $\times$ B: $\tfrac15(1\cdot4+2\cdot(-1)+2\cdot(-1)) = 0$; $\tfrac15(1\cdot(-2)+2\cdot3+2\cdot(-2)) = 0$; $\tfrac15(1\cdot(-1)+2\cdot(-1)+2\cdot4)=\tfrac{5}{5}=1$.

Thus $AB = \begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix} = I$. Since $AB = I$, $B = A^{-1}$; the matrices are inverses of each other.

Part (b): Prove $2\alpha+\beta = \dfrac{\pi}{4}$, given $\tan\alpha=\tfrac13,\ \tan\beta=\tfrac17$.

Step 1 — Find $\tan 2\alpha$ using $\tan 2\alpha = \dfrac{2\tan\alpha}{1-\tan^2\alpha}$: …